Fourier Transform Solved Problems

Introduction

Working through solved Fourier Transform problems builds the computational fluency needed for examinations and practical applications. This collection demonstrates the most important techniques: direct evaluation of the transform integral, strategic use of properties to avoid integration, inverse transformation, and system analysis in the frequency domain. Each solution emphasizes the reasoning process so you can apply similar strategies to new problems.

The key to solving Fourier Transform problems efficiently lies in recognizing which approach to use. Direct integration works for simple waveforms; properties save time when you can relate the given signal to a known pair; partial fractions handle rational spectra; and Parseval's theorem offers alternative routes for energy calculations. Study the method selection in each problem as carefully as the algebra.

Problem 1: Transform of a Decaying Exponential

Find $X(j\omega)$ for $x(t) = e^{-at}u(t)$ where $a > 0$.

Solution: Apply the Fourier Transform definition: $$X(j\omega) = \int_0^{\infty} e^{-at}e^{-j\omega t}dt = \int_0^{\infty}e^{-(a+j\omega)t}dt$$

Evaluating the integral: $$X(j\omega) = \left[\frac{e^{-(a+j\omega)t}}{-(a+j\omega)}\right]_0^{\infty} = 0 - \frac{1}{-(a+j\omega)} = \frac{1}{a+j\omega}$$

The convergence requires $a > 0$ (so the exponential decays as $t \to \infty$).

Magnitude: $|X(j\omega)| = \frac{1}{\sqrt{a^2+\omega^2}}$

Phase: $\angle X(j\omega) = -\arctan(\omega/a)$

This is a low-pass spectrum — magnitude decreases with frequency. The 3-dB bandwidth is at $\omega = a$ rad/s, where the magnitude drops to $1/\sqrt{2}$ of its DC value.

Physical insight: A faster decay (larger $a$) means the signal is shorter in time, so its spectrum is wider (bandwidth = $a$). This confirms the time-bandwidth tradeoff.

Problem 2: Transform Using Properties

Find $\mathcal{F}\{te^{-at}u(t)\}$ for $a > 0$.

Solution: Use the differentiation-in-frequency property: if $\mathcal{F}\{x(t)\} = X(j\omega)$, then $\mathcal{F}\{tx(t)\} = j\frac{dX}{d\omega}$.

We know $\mathcal{F}\{e^{-at}u(t)\} = \frac{1}{a+j\omega}$.

Therefore: $$\mathcal{F}\{te^{-at}u(t)\} = j\frac{d}{d\omega}\left(\frac{1}{a+j\omega}\right) = j \cdot \frac{-j}{(a+j\omega)^2} = \frac{1}{(a+j\omega)^2}$$

Extension: By applying the property repeatedly, we can show: $$\mathcal{F}\{t^n e^{-at}u(t)\} = \frac{n!}{(a+j\omega)^{n+1}}$$

This generates an entire family of transform pairs from a single known result, demonstrating the power of properties over brute-force integration.

Problem 3: Inverse Fourier Transform

Find $x(t)$ for $X(j\omega) = \frac{2j\omega}{(j\omega)^2 + 3j\omega + 2}$.

Solution: Factor the denominator: $(j\omega+1)(j\omega+2)$.

Partial fractions: $$\frac{2j\omega}{(j\omega+1)(j\omega+2)} = \frac{A}{j\omega+1} + \frac{B}{j\omega+2}$$

To find $A$: multiply both sides by $(j\omega+1)$ and set $j\omega = -1$: $$A = \frac{2(-1)}{-1+2} = -2$$

To find $B$: multiply both sides by $(j\omega+2)$ and set $j\omega = -2$: $$B = \frac{2(-2)}{-2+1} = 4$$

$$X(j\omega) = \frac{-2}{j\omega+1} + \frac{4}{j\omega+2}$$

Using the known pair $\frac{1}{a+j\omega} \leftrightarrow e^{-at}u(t)$:

$$x(t) = (-2e^{-t} + 4e^{-2t})u(t)$$

Verification: At $t=0$: $x(0) = -2 + 4 = 2$. From the initial value theorem: $x(0^+) = \lim_{\omega\to\infty}j\omega X(j\omega) = \lim_{\omega\to\infty}\frac{2(j\omega)^2}{(j\omega)^2+3j\omega+2} = 2$ ✓

Problem 4: Energy Calculation via Parseval's Theorem

Find the energy of $x(t) = e^{-2t}u(t)$ using the frequency domain.

Solution: $X(j\omega) = \frac{1}{2+j\omega}$

$$E = \frac{1}{2\pi}\int_{-\infty}^{\infty}|X(j\omega)|^2 d\omega = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{4+\omega^2}d\omega$$

Using the standard integral $\int_{-\infty}^{\infty}\frac{1}{a^2+\omega^2}d\omega = \frac{\pi}{a}$:

$$E = \frac{1}{2\pi} \cdot \frac{\pi}{2} = \frac{1}{4}$$

Verification (time domain): $E = \int_0^{\infty}e^{-4t}dt = \left[\frac{e^{-4t}}{-4}\right]_0^{\infty} = \frac{1}{4}$ ✓

When to use Parseval: If the time-domain integral is harder than the frequency-domain integral (or vice versa), choose whichever is simpler. For rational spectra, the frequency-domain route often involves standard integral formulas.

Problem 5: System Output in Frequency Domain

Given: Input $x(t) = e^{-t}u(t)$ passes through a system with $H(j\omega) = \frac{1}{1+j\omega}$.

Find the output $y(t)$.

Solution: $X(j\omega) = \frac{1}{1+j\omega}$

By the convolution theorem: $Y(j\omega) = X(j\omega) \cdot H(j\omega) = \frac{1}{(1+j\omega)^2}$

From the known pair: $\frac{1}{(a+j\omega)^2} \leftrightarrow te^{-at}u(t)$

$$y(t) = te^{-t}u(t)$$

Observation: When the input and system have the same pole ($s = -1$), the output has a repeated pole, producing a $te^{-t}$ response rather than simple exponentials. This is the frequency-domain manifestation of resonance between the input and system natural frequency.

Problem 6: Bandwidth Determination

Find the 3-dB bandwidth of the system $H(j\omega) = \frac{10}{5 + j\omega}$.

Solution: $|H(j\omega)| = \frac{10}{\sqrt{25+\omega^2}}$

At DC: $|H(0)| = 10/5 = 2$

At the 3-dB point, the magnitude drops to $1/\sqrt{2}$ of its maximum: $$|H(\omega_{3dB})| = \frac{2}{\sqrt{2}}$$ $$\frac{10}{\sqrt{25+\omega_{3dB}^2}} = \frac{2}{\sqrt{2}} \implies 25 + \omega_{3dB}^2 = 50 \implies \omega_{3dB} = 5 \text{ rad/s}$$

Bandwidth in Hertz: $f_{3dB} = 5/(2\pi) \approx 0.796$ Hz.

General rule: For a first-order system $H(j\omega) = \frac{K}{a+j\omega}$, the 3-dB bandwidth always equals $a$ rad/s. This is the system's time constant inverse — faster systems have wider bandwidths.

Problem 7: Transform of a Rectangular Pulse

Find $X(j\omega)$ for $x(t) = A\,\text{rect}(t/T)$ (amplitude $A$, width $T$, centered at origin).

Solution: By definition: $$X(j\omega) = \int_{-T/2}^{T/2} A\,e^{-j\omega t}\,dt = A\left[\frac{e^{-j\omega t}}{-j\omega}\right]_{-T/2}^{T/2}$$

$$= \frac{A}{-j\omega}\left(e^{-j\omega T/2} - e^{j\omega T/2}\right) = \frac{A}{-j\omega}(-2j\sin(\omega T/2))$$

$$= \frac{2A\sin(\omega T/2)}{\omega} = AT\,\text{sinc}\left(\frac{\omega T}{2\pi}\right)$$

Key observations:

• The spectrum is real (because $x(t)$ is real and even)
• First zero crossing at $\omega = 2\pi/T$ (narrower pulse → wider spectrum)
• Peak value $X(0) = AT$ (area under the pulse)
• The sinc function has sidelobes that decay as $1/\omega$

Problem 8: Frequency-Domain Differentiation

Find the Fourier Transform of $x(t) = \frac{t}{(1+t^2)^2}$.

Solution: Recognize that $\frac{d}{dt}\left(\frac{1}{1+t^2}\right) = \frac{-2t}{(1+t^2)^2}$

So $x(t) = -\frac{1}{2}\frac{d}{dt}\left(\frac{1}{1+t^2}\right)$

We know $\frac{1}{1+t^2} \xleftrightarrow{\mathcal{F}} \pi e^{-|\omega|}$ (from duality or direct computation).

Applying the time-differentiation property $\frac{dx}{dt} \leftrightarrow j\omega X(j\omega)$:

$$\mathcal{F}\left\{\frac{d}{dt}\frac{1}{1+t^2}\right\} = j\omega \cdot \pi e^{-|\omega|}$$

Therefore: $$X(j\omega) = -\frac{1}{2} \cdot j\omega\pi e^{-|\omega|} = -\frac{j\pi\omega}{2}e^{-|\omega|}$$

This demonstrates how combining known pairs with properties avoids a difficult direct integration.

Problem-Solving Strategy

1. Direct integration: Use for simple signals (exponentials, rectangles, impulses) where the integral is straightforward
2. Properties: Use scaling, shifting, differentiation, and duality to relate new signals to known pairs
3. Transform tables: Memorize the 8-10 key pairs and derive all others using properties
4. Partial fractions: Essential for inverse transforms of rational functions — factor the denominator, expand, and use known pairs
5. Parseval's theorem: Use for energy/power calculations when the frequency-domain form is simpler
6. Verification: Use initial/final value theorems, energy conservation, or known special values to check your answer

Key Takeaways

• For $e^{-at}u(t)$: $\mathcal{F} = 1/(a+j\omega)$ — the most fundamental transform pair
• Use the differentiation property to handle $t^n e^{-at}u(t)$ without integration
• Partial fractions + known pairs handle all rational Fourier Transforms
• Parseval's theorem provides an alternative route for energy computation
• System output: $Y = H \cdot X$ in frequency domain, then inverse transform
• Always verify results using independent methods when possible
• The choice of solution method is as important as the algebra itself